let $(M,g)$ be a Riemannian Manifold, let $ \gamma : [a,b] \rightarrow M$ be a piecewise smooth curve and let $\Omega : M \rightarrow \mathbb{R}^{n}$ be a coordinate chart. The length of $\gamma$ on the manifold is given by
$$L_{g}(\gamma) = \int_{a}^{b} | \gamma'(t)|_{g} dt.$$
$\gamma'(t)$ is an element of the tangent space to M at $\gamma(t)$. In local coordinates it is given by the equation
$$\gamma'(t) = (\gamma^{i})'(t) \frac{\partial}{\partial x^{i}} | _{\gamma(t)}$$
where $\gamma^{i}$ is $\gamma$ composed with the coordinate chart composed with the ith projection map. By definition
$$| \gamma'(t)|_{g} = (g_{\gamma(t)}(\gamma'(t),\gamma'(t)))^{\frac{1}{2}} =[ g_{ij} (\gamma^{i})'(t) (\gamma^{j})'(t) ]^{\frac{1}{2}} $$
Where $g_{ij}$ is the $(i,j)$ component of the metric tensor. You then just plug this into the integral and evaluate (with a computer I hope).
Now to calculate the distance between two points $a$ and $b$ on $M$, just calculate the length of a geodesic beginning at $a$ and ending at $b$. (This can be done on a sphere as geodesics are great circles. I am not so sure about an arbitrary riemannian manifold though...)